Bivariable differential equation of first order I found on an exercise sheet in physics the integration of the equation:
$$2 \frac{dX_0}{d\tau} = - \frac{X_0 (\tau)}{C_L(\tau)} \frac{dC_L}{d\tau} - C_L(\tau)$$
This equation can be thought as the following bivariable problem:
$$2 f'(x) = - \frac{f(x)}{g(x)} g'(x) - g(x)$$
I've looked at Wolfram Alpha, and at integro-differential sheets and I didn't find solution methods for solving this. Yet, the physics book seemed to imply that it was easy (maybe uninteresting?).
The solution they found is:
$$X_0(\tau) = \left( E - \frac{C_0}{2} \int_{1}^{\eta(1)} \sqrt{\frac{C_L(\tau)}{C_0}}  d\tau\right) \sqrt{\frac{C_L(\tau)}{C_0}}$$
Rephrased in our formulation:
$$f(x) = \left( E - \frac{g(0)}{2} \int_{1}^{\eta(1)} \sqrt{\frac{g(x)}{g(0)}}  d\tau\right) \sqrt{\frac{g(x)}{g(0)}}$$
$E$ is the boundary condition $X=E$ where $\eta = 1$.
One can rephrase the problem with this form:
$$2 \left( log(f(x)) \right)'(x) = -  \left( log(g(x)) \right)'(x) - \frac{g(x)}{f(x)}$$
I would like to know a method, or general methods to solve that kind of equations or this specific equation.
Thanks in advance.
 A: If $\;f(x)\not=0,\;$ then
$$2gff'+f^2g'+g^2f=0,$$
$$(f^2g)' = -g^2f,\tag1$$
$$-\dfrac{(f^2g)'}{f^4g^2} = \dfrac1{f^3},$$
$$\dfrac1{f^2g} = \int\dfrac1{f^3}\,\text dx,$$
$$\color{green}{\mathbf{g(x)=\dfrac1{f^2(x)\int\dfrac1{f^3(x)}\,\text dx}.\tag2}}$$
In the general case, expression in $RHS(2)$ for each given function $\;f(x)\;$ allows to define corresponding function $\;g(x),\;$ if it exists.
The alternative way.
Let us start from $(1)$ under the conditions
$$f(1)=E>0,\quad f(x)>0,\quad g(x)\not=0.\tag3$$
From $(3)$ follows that both f and g does not change the signs.
If $\;\mathbf{g(x)>0},\;$ then
$$\dfrac12\dfrac{(f^2g)'}{\sqrt{f^2g\,}} = -\dfrac12g^{\large\,^3/_2},$$
$$f(x)\,\sqrt{g(x)}=C-\dfrac12\int\limits_1^xg^{\large\,^3/_2}(t)\,\text dt,\quad C=E\,\sqrt{g(1)},$$
$$f(x)=\dfrac1{\sqrt{g(x)}}\left(E\,\sqrt{g(1)}-\dfrac12\int\limits_1^xg^{\large\,^3/_2}(t)\,\text dt\right).\tag{4a}$$
If $\;\mathbf{g(x)<0},\;$ then
$$\dfrac12\dfrac{(-f^2g)'}{\sqrt{-f^2g\,}} = \dfrac12(-g) ^{\large\,^3/_2},$$
$$f(x)\,\sqrt{-g(x)}=C+\dfrac12\int\limits_1^x(-g(t)) ^{\large\,^3/_2}\,\text dt,\quad C=E\,\sqrt{-g(1)},$$
$$f(x)=\dfrac1{\sqrt{-g(x)}} \left(E\,
\sqrt{-g(1)}+\dfrac12\int\limits_1^x(-g(t)) ^{\large\,^3/_2}\,\text dt\right).\tag{4b}$$
Finally,
$$\color{green}{\mathbf{f(x)=\dfrac1{\sqrt{|g(x)|}}\left(E\,\sqrt{|g(1)|}-\dfrac12 \operatorname{sgn}(g(1))  \int\limits_1^x|g(t)|^{\large\,^3/_2}\,\text dt\right).\tag5}}$$
Testing.
Let $\;g(x)=e^{-2x},\;$ then
$$f(x)=e^x\left(E-\dfrac12\int\limits_1^x e^{-3t}\,\text dt\right)
=\dfrac16e^x\left(6E-e^{-3}+e^{-3x}\right),$$
$$2f'(x)=\dfrac13e^x\left(6E-e^{-3}-2e^{-3x}\right),$$
$$f(x)\dfrac{g'(x)}{g(x)}=-\dfrac13e^x\left(6E-e^{-3}+e^{-3x}\right),$$
$$2f'(x)+f(x)\dfrac{g'(x)}{g(x)}+g(x)=0,$$
$$f(1)=E.$$
Testing confirms $(5).$
